Proof of Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability

Throughout, the values of a finite measure are real numbers $\ge0$, on which the sum and the product of $[0,\infty]$ are those of $\mathbb{R}$; two functions on $\mathcal{F}$ are equal when their values agree at every $A\in\mathcal{F}$; and $\mathcal{F}$ is closed under the set operations used below (routine facts and background of the setting).

Step 1: the Jordan form; a finite measure. By the Hahn decomposition theorem (which uses the axiom of dependent choice), every finite signed measure on $(X,\mathcal{F})$ has a Hahn set; in particular the Jordan decompositions used below exist. Let $P$ be a Hahn set for $\nu$ and $A\in\mathcal{F}$. By Total Variation and Total Variation Norm of a Finite Signed Measure (i) and Existence of the Total Variation of a Finite Signed Measure (i),

$$|\nu|(A)=\nu(A\cap P)-\nu(A\setminus P)=\nu^{+}(A)+\nu^{-}(A).$$

As $\nu^{+}(A)\ge0$ and $\nu^{-}(A)\ge0$, both are at most $|\nu|(A)$; this proves (ii). Thus $|\nu|=\nu^{+}+\nu^{-}$ setwise, a finite measure by Nonnegative Combinations of Finite Measures and Their Integrals (i) with $a_{1}=a_{2}=1$, which proves (i).

Step 2: the bound. Let $A\in\mathcal{F}$. Taking $B=A$ and $B=\emptyset$ in Total Variation and Total Variation Norm of a Finite Signed Measure (i), and using $\nu(\emptyset)=0$, both $\nu(A)$ and $-\nu(A)$ are at most $|\nu|(A)$, so $|\nu(A)|\le|\nu|(A)$. Since $|\nu|$ is a measure by Step 1, it is monotone, and $|\nu|(A)\le|\nu|(X)=\|\nu\|$. This proves (iii).

Step 3: nonnegative signed measures. Suppose $\nu(A)\ge0$ for every $A\in\mathcal{F}$. Every measurable subset of $X$ has nonnegative measure, and the only subset of $X\setminus X=\emptyset$ is $\emptyset$, of measure $0$; so $X$ is a Hahn set for $\nu$. Forming the Jordan decomposition with $P=X$,

$$\nu^{+}(A)=\nu(A\cap X)=\nu(A),\qquad\nu^{-}(A)=-\nu(A\setminus X)=-\nu(\emptyset)=0\qquad(A\in\mathcal{F}),$$

and Step 1 gives $|\nu|(A)=\nu(A)$. Hence $\nu=|\nu|=\nu^{+}$ is a finite measure by Step 1. Moreover $\nu^{-}$ is the zero measure, so $\int g\,d\nu^{-}=0$ for every $g\in\mathcal{M}^{+}$. Let $f\in\mathcal{L}^{1}(\nu)$. Since $\nu^{+}=\nu$, $f\in\mathcal{L}^{1}(\nu^{+})$ with $\int f\,d\nu^{+}=\int f\,d\nu$; and $\int f^{+}\,d\nu^{-}=\int f^{-}\,d\nu^{-}=0$, so $f\in\mathcal{L}^{1}(\nu^{-})$ with $\int f\,d\nu^{-}=0$. This proves (iv).

Step 4: linear combinations. Let $\nu'^{+}$ and $\nu'^{-}$ be the Jordan decomposition of $\nu'$. For every $A\in\mathcal{F}$,

$$(\nu+\nu')(A)=\bigl(\nu^{+}(A)+\nu'^{+}(A)\bigr)-\bigl(\nu^{-}(A)+\nu'^{-}(A)\bigr),$$

and $\nu^{+}+\nu'^{+}$ and $\nu^{-}+\nu'^{-}$ are finite measures by Nonnegative Combinations of Finite Measures and Their Integrals (i); a difference of two finite measures is a finite signed measure, so $\nu+\nu'$ is one. Likewise $(c\nu)(A)=c\,\nu^{+}(A)-c\,\nu^{-}(A)$ if $c\ge0$, and $(c\nu)(A)=|c|\,\nu^{-}(A)-|c|\,\nu^{+}(A)$ if $c<0$, where $c\nu^{\pm}$ and $|c|\nu^{\pm}$ are finite measures by Nonnegative Combinations of Finite Measures and Their Integrals (i) with $\mu_{2}=\mu_{1}$ and $a_{2}=0$. This proves (v).

Step 5: total variation of linear combinations. Let $A\in\mathcal{F}$; by Step 4 the total variations of $\nu+\nu'$ and $c\nu$ are defined. For $B\in\mathcal{F}$ with $B\subseteq A$,

$$(\nu+\nu')(B)-(\nu+\nu')(A\setminus B)=\bigl(\nu(B)-\nu(A\setminus B)\bigr)+\bigl(\nu'(B)-\nu'(A\setminus B)\bigr)\le|\nu|(A)+|\nu'|(A),$$

so the greatest of these numbers, $|\nu+\nu'|(A)$, is at most $|\nu|(A)+|\nu'|(A)$. Next, $(c\nu)(B)-(c\nu)(A\setminus B)=c\bigl(\nu(B)-\nu(A\setminus B)\bigr)$. If $c\ge0$, the set of these numbers is $\{ct:t\in S_{\nu}(A)\}$, with $S_{\nu}(A)$ as in Existence of the Total Variation of a Finite Signed Measure (i), and since $t\mapsto ct$ is nondecreasing its greatest element is $c\,|\nu|(A)$. If $c<0$, put $B'=A\setminus B$, a measurable subset of $A$ with $A\setminus B'=B$; then

$$c\bigl(\nu(B)-\nu(A\setminus B)\bigr)=|c|\bigl(\nu(B')-\nu(A\setminus B')\bigr),$$

and as $B$ runs through the measurable subsets of $A$ so does $B'$, the function $B\mapsto A\setminus B$ being its own inverse on them. So the set of these numbers is $\{|c|t:t\in S_{\nu}(A)\}$, with greatest element $|c|\,|\nu|(A)$. This proves (vi).

Step 6: the normed space. $M(X,\mathcal{F})$ is the subset of $\mathbb{R}^{\mathcal{F}}$ given by the condition of being a finite signed measure, a set by the routine facts of the setting. The zero function $z$ on $\mathcal{F}$ lies in it: $z(\emptyset)=0$, and for every sequence the series of zeros converges with sum $0$. By Step 4 the sum is a binary operation on $M(X,\mathcal{F})$ and the scalar multiplication a function $\mathbb{R}\times M(X,\mathcal{F})\to M(X,\mathcal{F})$. Each law of a real vector space not involving the zero or negatives, and the identity $\nu+z=\nu$, reduces at every $A\in\mathcal{F}$ to the corresponding law of the field $\mathbb{R}$; so Vector Space over a Field (iii) holds with $z$. For Vector Space over a Field (iv), let $v\in M(X,\mathcal{F})$ and let $w\in M(X,\mathcal{F})$ satisfy $\rho+w=\rho$ for every $\rho\in M(X,\mathcal{F})$; taking $\rho=z$ gives $w=z$, and $y=(-1)v$ lies in $M(X,\mathcal{F})$ with $v+y=z$. Hence $M(X,\mathcal{F})$ is a real vector space, and $z$ is its zero vector.

We check the conditions of Norms and Real Normed Spaces (v). Nonnegativity: $\|\nu\|=|\nu|(X)\ge0$, $|\nu|$ being a measure. Definiteness: if $\|\nu\|=0$, then $|\nu(A)|\le0$ for every $A\in\mathcal{F}$ by Step 2, so $\nu=z$. Absolute homogeneity and the triangle inequality are Step 5 with $A=X$. Hence $\|\cdot\|$ is a norm on $M(X,\mathcal{F})$. Finally $\|z\|=\|0z\|=0\,\|z\|=0$, which with definiteness gives the last assertion. This proves (vii).

Step 7: integrability. Let $f:X\to\mathbb{R}$ be measurable. By the integrability criterion of the base, $|f|\in\mathcal{M}^{+}$, and for every measure $\rho$ on $(X,\mathcal{F})$, $f\in\mathcal{L}^{1}(\rho)$ if and only if $\int|f|\,d\rho<\infty$. Since $|\nu|=\nu^{+}+\nu^{-}$ setwise by Step 1, Nonnegative Combinations of Finite Measures and Their Integrals (ii) with $a_{1}=a_{2}=1$ and $g=|f|$ gives

$$\int|f|\,d|\nu|=\int|f|\,d\nu^{+}+\int|f|\,d\nu^{-}.$$

A sum of two elements of $[0,\infty]$ is finite exactly when both are, so $f\in\mathcal{L}^{1}(|\nu|)$ if and only if $f\in\mathcal{L}^{1}(\nu^{+})$ and $f\in\mathcal{L}^{1}(\nu^{-})$. This proves (viii).